Adiabatic particle pumping and anomalous...

Adiabatic particle pumping and anomalous velocity

November 17, 2015

November 17, 2015 1 / 31

Literature:1 J. K. Asboth, L. Oroszlany, and A. Palyi, arXiv:1509.02295

2 D. Xiao, M-Ch Chang, and Q. Niu, Rev. Mod. Phys. 82, 1959.

November 17, 2015 2 / 31

SSH model, adiabatic evolution, Chern number

A couple of the ideas discussed on previous occasions will be combined.

i) Su-Schrieffer-Heeger (Rice-Mele) model

ii) Adiabatic evolution of eigenstates

iii) Chern number

November 17, 2015 3 / 31

SSH model, adiabatic evolution, Chern number

A couple of the ideas discussed on previous occasions will be combined.

i) Su-Schrieffer-Heeger (Rice-Mele) model

ii) Adiabatic evolution of eigenstates

iii) Chern number

Figure : A segment of the periodic SSH model.

November 17, 2015 3 / 31

Overview

1) Model system

2) Particle current and group velocity

3) Quasi adiabatic time evolution

4) Pumped current and Berry curvature

5) Anomalous velocity of electrons

November 17, 2015 4 / 31

Model system

Because of spatial periodicity, the eigenstates are Bloch wave functions.The instantaneous eigenstates read |Ψn,k(t)〉 = |k〉 ⊗ |un(k , t)〉, wheren = 1, 2 are the two bands, k wavenumber.

November 17, 2015 6 / 31

Model system

Because of spatial periodicity, the eigenstates are Bloch wave functions.The instantaneous eigenstates read |Ψn,k(t)〉 = |k〉 ⊗ |un(k , t)〉, wheren = 1, 2 are the two bands, k wavenumber.Reminder:

|k〉 = 1√N

N∑

m=1

e imk |m〉 m : site index

and |un(k , t)〉 satisfy the instantaneous Schrodinger equation

H(k , t)|un(k , t)〉 = En(k , t)|un(k , t)〉.

November 17, 2015 6 / 31

Model system


|k〉 = 1√N

N∑

m=1




Note, that this is a different convention regarding |un(k , t)〉 than whatwe used the last time to define the Bloch wave functions.

November 17, 2015 6 / 31

Model system


|k〉 = 1√N

N∑

m=1




Note, that this is a different convention regarding |un(k , t)〉 than whatwe used the last time to define the Bloch wave functions.

Nevertheless, it is true that |Ψn,k(r + Rl)〉 = e ikRl |Ψn,k(r)〉

November 17, 2015 6 / 31

Model system

We assume that the Hamiltonian is also periodic in time, therefore

1 H(k + 2π, t) = H(k , t)

2 H(k , t + T ) = H(k , t), and Ω = 2π/T .

November 17, 2015 7 / 31

Model system

We assume that the Hamiltonian is also periodic in time, therefore

1 H(k + 2π, t) = H(k , t)

2 H(k , t + T ) = H(k , t), and Ω = 2π/T .

We can rewrite the problem using the following notation. The Hamiltonianis given by (two-band insulator model):

H(d(k , t)) = dxσx + dyσy + dzσz = d · σThe parameter vector d(k , t) reads:

d(k , t) =

ν + cosΩt + cos ksin ksinΩt

⇐⇒v(t) = ν + cosΩt

w(t) = 1u(t) = sinΩt

November 17, 2015 7 / 31

Overview

1) Model system





November 17, 2015 8 / 31

Influx of particles into a segment


The number of particles NS in segment S is given by the expectationvalue of the operator

NS =∑

m∈S

∑

α∈A,B

|m, α〉〈m, α|

November 17, 2015 9 / 31



The number of particles NS in segment S is given by the expectationvalue of the operator

NS =∑

m∈S

∑

α∈A,B

|m, α〉〈m, α|

Due to the time-dependence, the number of particles in a given regionchanges.

November 17, 2015 9 / 31



The change of the number of particles is given by the Heisenberg equationof motion

∂N (t)S∂t

= jS (t) = −i [N , H]

November 17, 2015 10 / 31

Particle current operator



The particle number can change through the interface at m = p andthrough the interface at m = q.

November 17, 2015 11 / 31

Particle current operator



The particle number can change through the interface at m = p andthrough the interface at m = q.One can define

jm+1/2(t) = −iw(t) (|m + 1,A〉〈m,B | − |m,B〉〈m + 1,A|)which is interpreted as the operator describing the net particle flow

through the cross-section at m + 1/2.November 17, 2015 11 / 31


jm+1/2(k , t) =1

N

(

0 −iw(t)e−ik

iw(t)e ik 0

)

We can recognize that

jm+1/2(k , t) =1

N∂k H(d(k , t))

=⇒ the momentum diagonal elements of the current operator are relatedto the momentum-space Hamiltonian.

November 17, 2015 13 / 31


jm+1/2(k , t) =1

N

(

0 −iw(t)e−ik

iw(t)e ik 0

)

We can recognize that

jm+1/2(k , t) =1

N∂k H(d(k , t))

=⇒ the momentum diagonal elements of the current operator are relatedto the momentum-space Hamiltonian.

This is a very general result. It applies not only to the Rice-Mele problembut to time-independent problems defined on a lattice and can begeneralized to (quasi) two dimensional case (Buttiker formalism etc.)

November 17, 2015 13 / 31


Matrix elements of the current operator: in terms of the instantaneouseigenvalues

〈un(k , t)|jm+1/2(k , t)|un(k , t)〉 =1

N〈un(k , t)|∂k H|un(k , t)〉

=∂kE (k , t)

N=

vn(k , t)

N

Here vn(k , t) is the instantaneous group velocity of the eigenstate.

November 17, 2015 14 / 31

Overview

1) Model system





November 17, 2015 15 / 31

Time evolution governed by quasi-adiabatic Hamiltonian

We want to calculate the total particle current over one cycle of driving.

November 17, 2015 16 / 31



For this we need to consider the time evolution of the eigenstates (notonly the instantaneous eigenstates)

November 17, 2015 16 / 31



For this we need to consider the time evolution of the eigenstates (notonly the instantaneous eigenstates)

Remarks

i) due to the spatial periodicity, the wave number k remains a goodquantum number. Therefore we consider the time evolution of a set oftwo-level system, each labelled by a different k

ii) basically, we are going to use time dependent perturbation theory tocalculate the time evolution of the eigenstates

November 17, 2015 16 / 31


Quasi adiabatic: for the driving Ω it is fulfilled that Ω ≪ ∆E where∆E = min∀k,t,t→T (E2(k , t)− E1(k , t)).

November 17, 2015 17 / 31


Quasi adiabatic: for the driving Ω it is fulfilled that Ω ≪ ∆E where∆E = min∀k,t,t→T (E2(k , t)− E1(k , t)).We are looking for the solutions of the following time dependentSchrodinger equation.

i~∂

∂t|u(t)〉 = H(t)|u(t)〉

One can expand the wave function in terms of the instantaneouseigenstates of H(t)

|u(t)〉 =∑

n

exp

(

− i

~

∫ t

0dt ′En(t

′)

)

an(t)|un(t)〉

November 17, 2015 17 / 31


Quasi adiabatic: for the driving Ω it is fulfilled that Ω ≪ ∆E where∆E = min∀k,t,t→T (E2(k , t)− E1(k , t)).We are looking for the solutions of the following time dependentSchrodinger equation.

i~∂

∂t|u(t)〉 = H(t)|u(t)〉

One can expand the wave function in terms of the instantaneouseigenstates of H(t)

|u(t)〉 =∑

n

exp

(

− i

~

∫ t

0dt ′En(t

′)

)

an(t)|un(t)〉

The coefficients satisfy

∂tan(t) = −∑

l

al(t)〈un(t)|∂t |ul(t)〉 exp(

− i

~

∫ t

0dt ′[El (t

′)− En(t′)]

)

November 17, 2015 17 / 31



〈un(t)|∂t |un(t)〉 = ∂td(t)〈un(t)|∂

∂d|un(t)〉 = 0.


This condition basically corresponds to neglecting the acquired Berryphase, which turns out to be not important in this case.

November 17, 2015 18 / 31



〈un(t)|∂t |un(t)〉 = ∂td(t)〈un(t)|∂

∂d|un(t)〉 = 0.



The characteristic frequency for ∂td(t) is Ω.If ∂td(t) → 0 then in zeroth order

∂tan(t) = 0

November 17, 2015 18 / 31



〈un(t)|∂t |un(t)〉 = ∂td(t)〈un(t)|∂

∂d|un(t)〉 = 0.



The characteristic frequency for ∂td(t) is Ω.If ∂td(t) → 0 then in zeroth order

∂tan(t) = 0

If the system is initially in the nth eigenstate then approximately it willstay in that eigenstate =⇒ adiabatic theorem.

November 17, 2015 18 / 31


First order corrections:Initially, an(t = 0) = 1 and an′ 6=n(t = 0) = 0.The equation for an′ 6=n:

∂tan′ = −〈un′(t)|∂t |un(t)〉 exp(

− i

~

∫ t

0dt ′[En(t

′)− En′(t′)]

)

November 17, 2015 19 / 31


First order corrections:Initially, an(t = 0) = 1 and an′ 6=n(t = 0) = 0.The equation for an′ 6=n:

∂tan′ = −〈un′(t)|∂t |un(t)〉 exp(

− i

~

∫ t

0dt ′[En(t

′)− En′(t′)]

)

The approximate solution, up to first order in ∂td(t) is then

an′ = −〈un′(t)|∂t |un(t)〉En − En′

i~ exp

(

− i

~

∫ t

0dt ′[En(t

′)− En′(t′)]

)

November 17, 2015 19 / 31



|u1(t)〉 = e−i∫ t0 dt′E1(t

′)

[

|u1(t)〉+ i〈u2(t)|∂t |u1(t)〉E2(t)− E1(t)

|u2(t)〉]


Most of the weight is in the instantaneous ground state |u1(t)〉 with asmall admixture of the instantaneous excited state |u2(t)〉.

In the case of cyclic change of parameters, the final state |u1(t = T )〉 maydiffer from the initial state |u1(t = 0)〉 by a phase factor e iγ1 (Berryphase).

November 17, 2015 20 / 31

Overview

1) Model system





November 17, 2015 21 / 31

Pumped probability current and Berry curvature


We want to calculate the number of particles N through a cross section atm + 1/2 in a time interval [0,T ].

N =

∫ T

0dt

∑

k∈BZ

〈Ψ1(k , t)|jm+1/2(t)|Ψ1(k , t)〉

November 17, 2015 22 / 31



We want to calculate the number of particles N through a cross section atm + 1/2 in a time interval [0,T ].

N =

∫ T

0dt

∑

k∈BZ

〈Ψ1(k , t)|jm+1/2(t)|Ψ1(k , t)〉

Here Ψ1(k , t) ≈ |k〉 ⊗ |u1(k , t)〉

November 17, 2015 22 / 31


We can write

N =

∫ T

0dt

∑

k∈BZ

〈u1(k , t)|jm+1/2(k , t)|u1(k , t)〉

=1

N

∫ T

0dt

∑

k∈BZ

〈u1(k , t)|∂k H(k , t)|u1(k , t)〉

November 17, 2015 23 / 31


〈u1(k , t)|jm+1/2(k , t)|u1(k , t)〉 = v1(k , t) + i〈u1|∂k H |u2〉〈u2|∂t |u1〉

E2 − E1+ c .c

November 17, 2015 24 / 31



E2 − E1+ c .c

When sumed over the Brillouin zone, the contribution ∼ v1(k , t) willvanish.

November 17, 2015 24 / 31



E2 − E1+ c .c


Next, we use that

〈u1|∂kH |u2〉 = (E1 − E2)〈∂ku1|u2〉

November 17, 2015 24 / 31



E2 − E1+ c .c


Next, we use that

〈u1|∂kH |u2〉 = (E1 − E2)〈∂ku1|u2〉

therefore

−〈u1|∂k H |u2〉〈u2|∂t |u1〉E2 − E1

+ c .c = −i〈∂ku1|u2〉〈u2|∂t |u1〉+ c .c .

November 17, 2015 24 / 31





The last expression is exactly the Berry curvature Ω(1)(k , t) defined in theparameter space (k , t).

November 17, 2015 25 / 31





The last expression is exactly the Berry curvature Ω(1)(k , t) defined in theparameter space (k , t).

Finally, the number of pumped particles in a given driving cycle is

N =

∫ T

0dt

∫

BZ

dk

2πΩ(1)(k , t)

given by the integral of the Berry curvature.

November 17, 2015 25 / 31


Figure : Number of pumped particles as a function of time.

November 17, 2015 26 / 31


Figure : Number of pumped particles as a function of time.

Does it have to be an integer number ?

November 17, 2015 26 / 31

Overview

1) Model system





November 17, 2015 27 / 31

Anomalous velocity of electrons

Let us consider a crystal under the perturbation of a weak electric field E.We represent the electric field through a uniform vector potential A(t)which is time dependent.

November 17, 2015 28 / 31



Using minimal coupling theory, the Hamiltonian can be written

H(t) =[p+ eA]2

2me+ V (r)

V (r) lattice periodic potential.

November 17, 2015 28 / 31



Using minimal coupling theory, the Hamiltonian can be written

H(t) =[p+ eA]2

2me+ V (r)

V (r) lattice periodic potential.

Because of the translation invariance, the solutions are Bloch wavefunctions.The Hamiltonian that acts on the lattice periodic part of the Bloch wavefunctions can be written

H(k, t) = H(

k+e

~A(t)

)

November 17, 2015 28 / 31


Since A preserves translation invariance, k is a good quantum number andis a constant of motion. Therefore

d

dtq =

d

dt

(

k+e

~A(t)

)

=e

~

d

dtA(t) = −e

~E

November 17, 2015 29 / 31



d

dtq =

d

dt

(

k+e

~A(t)

)

=e

~

d

dtA(t) = −e

~E

One can define the average velocity in band n as a function of q:

v(n)(q) = 〈un(k, t)|∂qH |un(k, t)〉

November 17, 2015 29 / 31



d

dtq =

d

dt

(

k+e

~A(t)

)

=e

~

d

dtA(t) = −e

~E

One can define the average velocity in band n as a function of q:

v(n)(q) = 〈un(k, t)|∂qH |un(k, t)〉

Using

i) the results for the adiabatic evolution of |un(k, t)〉ii) that ∂/∂ki = ∂/∂qi and ∂/∂t = −(e/~)Ei∂/∂qi

November 17, 2015 29 / 31


One can write the average velocity as

v(n)(q) =∂En(q)

~∂q− e

~E×Ω(n)(q)

where Ω(n)(q) is the Berry curvature of the nth band

Ω(n)(q) = i〈∇qun(k)| × |∇qun(q)〉

November 17, 2015 30 / 31


One can write the average velocity as

v(n)(q) =∂En(q)

~∂q− e

~E×Ω(n)(q)

where Ω(n)(q) is the Berry curvature of the nth band

Ω(n)(q) = i〈∇qun(k)| × |∇qun(q)〉

One can see that there is an anomalous velocity component

e

~E×Ω(n)(q)

which is transverse to the electric field.

November 17, 2015 30 / 31


Symmetry considerations

The velocity v(n)(q) should obey certain symmetry constraints:

i) under time reversal, v(n) and q) change sign, while E is fixed

ii) under spatial inversion, v(n), q), E change sign

November 17, 2015 31 / 31






This requires that

i) if the unperturbed system has time reversal symmetry thenΩ(n)(−q) = −Ω(n)(q)

ii) if the unperturbed system has spatial inversion symmetry thenΩ(n)(−q) = Ω(n)(q)

November 17, 2015 31 / 31






This requires that

i) if the unperturbed system has time reversal symmetry thenΩ(n)(−q) = −Ω(n)(q)

ii) if the unperturbed system has spatial inversion symmetry thenΩ(n)(−q) = Ω(n)(q)

In crystals with simultaneous time-reversal and inversion symmetry theBerry curvature vanishes in the whole Brillouin zone.

November 17, 2015 31 / 31

Adiabatic particle pumping and anomalous...

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